Question

The acceleration due to gravity $\left(g\right)$ is determined by using simple of length $l=(100\pm 0.1)cm$. If the time period is $T=(2\pm 0.01)s$, find the maximum percentage error in the measurment of $g$.

Answer 1

Given that,

Time period $T=(2\pm 0.01)s$

Length $L=(100\pm 0.1)cm$

We know that,

$T=2\pi \sqrt{\frac{l}{g}}$

$T^2=4{\pi }^2\times \frac{l}{g}$

$g=\frac{4{\pi }^{2}l}{T^2}$

Now, the maximum percentage error

$\frac{\Delta g}{g}=\pm (\frac{\Delta l}{l}+2\frac{\Delta T}{T})$

$\frac{\Delta g}{g}=\pm (\frac{0.1}{100}+2\times \frac{0.01}{2})$

$\frac{\Delta g}{g}=\pm (\frac{0.1}{1\stackrel{`}{}00}+0.01)$

$\frac{\Delta g}{g}=\pm 0.0011$

$\Delta g=\pm 9.8\times \pm 0.0011$

$\Delta g=0.01078$

$\Delta g\times 100=0.0108\times 100$

$\Delta g\times 100=1.08$%

Hence, the maximum percentage error in the measurement of g is $1.08$%